Q. The maximum velocity to which a proton can be accelerated in a cyclotron of 10 MHz frequency and radius 50 cm is
A) $6.28\ \times 10^{8}\ m/s$
B) $3.14\ \times 10^{8}\ m/s$
C) $6.28\ \times 10^{7}\ m/s$
D) $3.14\ \times 10^{7}\ m/s$
Explanation
(d) The motion of proton is magnetic field will be circular.
$v = 2\pi rf$
$\ \ \ = 2 \times 3.14 \times 50 \times 10^{- 2} \times 10 \times 10^{6}$
$\ \ \ = 3.14 \times 10^{7}\ m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.